## Limits Approaching A Finite Value - Finite Limits

### Video List

A09   A10   A11   Substitution - 3 Examples

A12   factoring

A13   expanding

B07   common denominator

A04  factoring 1

A05  factoring 2

A06  factoring 3

A07  factoring 4

A08  factoring 5

B06  factoring 6

A14   example 1

A15   example 2

A16   A17   examples 3

### Finite Limits, Infinite Limits, Limits At Infinity . . . Terminology Explained

The use of the terms finite limits, infinite limits and limits at infinity are used differently in various books and your instructor may have their own idea of what they mean. In this panel, we will try to break down the cases and explain the various ways these terms can be used as well as how we use them here at 17calculus.

When we talk about limits, we are looking at the $$\displaystyle{ \lim_{x \to c}{f(x)} = L }$$. The various terms apply to the description of $$c$$ and $$L$$ and are shown in the table below. The confusion lies with the terms finite limits and infinite limits. They can mean two different things.

$$\displaystyle{ \lim_{x \to c}{f(x)} = L }$$

when

term(s) used

$$c$$ is finite

limits approaching a finite value or finite limits

$$c$$ is infinite $$\pm \infty$$

limits at infinity or infinite limits

$$L$$ is finite

finite limits

$$L$$ is infinite $$\pm \infty$$

infinite limits

You can see where the confusion lies. The terms finite limits and infinite limits are used to mean two different things, referring to either $$c$$ or $$L$$. It is possible to have $$c = \infty$$ and $$L$$ be finite. So is this an infinite limit or a finite limit? It depends if you are talking about $$c$$ or $$L$$.

How 17calculus Uses These Terms
The pages on this site are constructed based on what $$c$$ is, i.e. we use the terms finite limits and infinite limits based on the value of $$c$$ only ( using the first two rows of the table above and ignoring the last two ). This seems to be the best way since, when we are given a problem, we can't tell what $$L$$ is until we finish the problem, and therefore we are unable to determine what type of problem we have and know what techniques to use.

Important: Make sure and check with your instructor to see how they use these terms.

Limits discussed on this page are approaching a constant, i.e. a specific finite number. In limit notation, they look like $$\displaystyle{ \lim_{x \rightarrow c}{~f(x)} }$$ where $$c$$ is a finite number.
If you want to study the case where $$c$$ is infinite, you need to go to the Limits At Infinity ( Infinite Limits ) page. ( The panel above explains the terminology and how 17calculus defines finite and infinite limits. )

Studying limits that approach a constant is the best place to start to get an understanding of how limits work. Make sure you understand the Limit Key before going on.

General Steps To Evaluate Limits

 Step 1 [ substitution ] - - To evaluate a limit at a constant, the first thing you will always want to try is direct substitution. From the Limit Key discussion, you know that a limit doesn't necessarily mean the value AT the point. However, if the function is continuous at the point in question, the limit will be equal to the function at that point. Plugging the number into the function will help you determine if that is the case. If you plug in the number into the function and you get a finite number, then you are done and that number is your answer. However, if you get $$0/0$$ (which is an indeterminate form meaning it could be anything) or a fraction with a non-zero number in the numerator and a zero in the denominator, you have a little more work to do. We need to look at each case separately. Step 2A ( indeterminate form ) - - Let's start with $$0/0$$. What's going here is that $$0/0$$ is what we call indeterminate. Basically, that means it can't be determined, i.e. $$0/0$$ can mean anything. It may mean $$0$$, it may mean $$27$$ or $$19075$$ or infinity. You can't tell. So what we need to do is to work some algebra on the problem to get it in a different form. Usually this means factoring the numerator and denominator to get a common factor that will cancel. [ See the videos below for demonstrations of this technique. ] Another option may to use trig identities so that factoring becomes possible. Step 2B ( zero denominator ) - - The other case is when you get a non-zero number in the numerator and zero in the denominator. This case means that the limit is either $$+\infty$$ or $$-\infty$$. In this case you need to look at the behavior of the denominator very near zero to see if it stays negative or positive on both sides of the limit. Step 3 - - Once you have altered the form of the limit, use substitution again to see if anything has changed. If not, then try again by going back to step 2 ( whichever applies ) and try again.

From the above list of steps, you can extrapolate four main techniques that you might used, depending on the situation. Below is a discussion of each technique in detail.

Individual Techniques

### Substitution

We know from the concept of continuity, that if a function, $$f(x)$$, is continuous at the point $$x=c$$, then $$\displaystyle{ \lim_{x \to c}{[f(x)]} = f(c) }$$. So this tells us, that if the function is continuous at the point at which we are taking the limit, then we can just substitute that point into the function to determine the limit and we are done.

We can also substitute the limit value into the function to help us determine what to do next, even if the function is not continuous at the point. Here is what substitution tells us.

Case 1: After substitution, if we get an indeterminate form, we need to use one (or more) of the techniques in the following panels (factoring, rationalizing, using trig identities or, if you know derivatives, L'Hôpital's Rule).

Case 2: After substitution, if we get a fraction like $$c/0$$ where $$c$$ is any non-zero finite number this indicates a possible asymptote, so we need to look at what is happening on either side of $$x=c$$. There are 4 possibilities.

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = +\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = +\infty }$$
The limit exists and $$\displaystyle{ \lim_{x \to c}{[f(x)]} = +\infty }$$.

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = -\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = -\infty }$$
The limit exists and $$\displaystyle{ \lim_{x \to c}{[f(x)]} = -\infty }$$.

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = +\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = -\infty }$$
or
$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = -\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = +\infty }$$
The limit does not exist.

Note
On this site (and in many textbooks), we say that when $$\displaystyle{ \lim_{x \to c}{[f(x)]} = \infty }$$, the limit exists and the limit is infinity (similarly for negative infinity). Not everyone agrees with this and your textbook and/or instructor may see this differently. So ask your instructor for clarification.

### Factoring

Once we have substituted the limit value in the function and get the indeterminate form $$0/0$$, we can try factoring and canceling to get a determinate form. This technique works best with polynomials or with terms that can factor.

When we get $$0/0$$, we have a hole in the graph (a removable discontinuity). In order to understand why this works, we invoke the following theorem.

#### Functions That Agree At All But One Point

Let $$c$$ be a real number and let $$g(x) = h(x)$$ for all $$x \neq c$$ in an open interval containing $$c$$.

If $$\displaystyle{ \lim_{x \to c}{[g(x)]} }$$ exists, then $$\displaystyle{ \lim_{x \to c}{[h(x)]} }$$ also exists and $$\displaystyle{ \lim_{x \to c}{[h(x)]} = \lim_{x \to c}{[g(x)]} }$$.

Proof

What this theorem says is that if we have two functions that are identical, expect (possibly) at $$x=c$$, and we know that one function has a limit that exists at $$c$$, then the other one is guaranteed to have a limit that exists at $$c$$ and that limit is the same as the limit of the first function.

This theorem is incredibly useful for functions with holes (our current topic). If we have a function with a hole, and if we can find a second function that is the same as first function, except at $$x=c$$, that has a limit there, then we can use that second function to determine the limit of the first function. Usually, this second function is simpler than the first and the way we find it is by factoring and canceling common terms.

Here is a quick example.
Let $$\displaystyle{ f(x) = \frac{x-1}{x^2-1} = \frac{x-1}{(x-1)(x+1)} }$$     and     $$\displaystyle{ g(x) = \frac{1}{x+1} }$$

Notice that by canceling the $$(x-1)$$ term in $$f(x)$$, we get $$g(x)$$. However, $$f(x) \neq g(x)$$ since the domains of the two functions are different. The domain of $$f(x)$$ is $$x \neq 1$$ and the domain of $$g(x)$$ is all real numbers.

However, we can say that $$\displaystyle{ \lim_{x \to 1}{[f(x)]} = \lim_{x \to 1}{[g(x)]} }$$ by applying the above theorem. Notice the difference, the functions are not equal but the limits at $$x=1$$ are equal. This is an important distinction.

See the practice problems below for more examples.

### Rationalizing

This is one of the more interesting techniques and one that you are extremely likely to need to use on an exam. The idea is that you are given a limit that is indeterminate containing a square root term. You rationalize either the numerator or denominator (whichever location has the square root) to remove the square root. This will usually lead to a factor that can then be canceled so that the result is no longer indeterminate.

This technique depends on the idea demonstrated in this example: $$(\sqrt{x} - 1)(\sqrt{x}+1) = x-1$$. Notice that the square root term is gone. In general, given a term that contains a square root, multiply by an identical term except for one sign change. For example, if you are given $$\sqrt{x+3} + 7$$, multiplying by $$\sqrt{x+3} - 7$$ will give you $$x+3 - 49 = x-46$$. It works the other way too, i.e. $$3 - \sqrt{3x}$$, multiplied by $$3 + \sqrt{3x}$$ yields $$9 - 3x$$.

Now, you can't just go around multiplying terms by whatever you want. In the problems where you will use this technique, you need to have the indeterminate form $$0/0$$ and you multiply the numerator and denominator by the same term, essentially multiplying the entire function by $$1$$. This is a valid operation that does not change the problem. Let's do a quick example because I want to show you something to watch out for.

Example
$$\displaystyle{ \begin{array}{rcl} \lim_{x \to 2}{\frac{\sqrt{x+2}-2}{x-2}} & = & \lim_{x \to 2}{\frac{\sqrt{x+2}-2}{x-2}}\frac{\sqrt{x+2}+2}{\sqrt{x+2}+2} \\ & = & \lim_{x \to 2}{\frac{x+2 - 4}{(x-2)(\sqrt{x+2}+2)}} \\ & = & \lim_{x \to 2}{\frac{x-2}{(x-2)(\sqrt{x+2}+2)}} \\ & = & \lim_{x \to 2}{\frac{1}{\sqrt{x+2}+2}} = 1/4 \end{array} }$$

Notes

1. We multiplied the numerator and denominator by the non-zero term $$\sqrt{x+2}+2$$. To get this, we took the term $$\sqrt{x+2}-2$$ and changed the sign between $$\sqrt{x+2}$$ and $$2$$, outside the square root. This essentially multiplied the entire function by $$1$$.
2. We did NOT multiply out the denominator when we had $$(x-2)(\sqrt{x+2}+2)$$. This will ALWAYS be the case, since the goal is to get a term that you can cancel, you want to leave the other (non-rationalized) term factored.
3. This technique also works when the term to be rationalized is in the denominator.
4. You may also use this technique when you have two square roots in a term. For example, $$\sqrt{x+3} - \sqrt{x-5}$$. The rationalization term will be $$\sqrt{x+3} + \sqrt{x-5}$$.

### Using Trig Identities and Special Trig Limits

In addition to using the standard trig identities to convert trig functions into other, more useful, forms, we can use these two special limits to determine the limit involving trig functions when we have an indeterminate form.

$$\displaystyle{ \lim_{\theta \to 0}{\frac{sin(\theta)}{\theta}} = 1 }$$     and     $$\displaystyle{ \lim_{\theta \to 0}{\frac{1-cos(\theta)}{\theta}} = 0 }$$

Among the identities that you will need are

$$cos^2(\theta) + sin^2(\theta) = 1$$

$$tan(\theta) = sin(\theta) / cos(\theta)$$

$$sec(\theta) = 1/cos(\theta)$$

$$csc(\theta) = 1/sin(\theta)$$

Detail discussion and practice problems can be found on the trig limits page.

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### Finite Limits Resources

use these filters to show only the practice problems you want to see

solve using substitution

solve by factoring

solve by rationalizing

Practice Problems

Instructions:
Unless otherwise instructed, evaluate the following limits, giving your answers in exact terms.

 Level A - Basic

Practice A01

$$\displaystyle{\lim_{x \rightarrow 2}{ \left[ \frac{3x^2-x-10}{x^2-4} \right] } }$$

Practice A02

$$\displaystyle{ \lim_{x \rightarrow 3}{ \left[\frac{x^4-81}{2x^2-5x-3}\right] } }$$

Practice A03

$$\displaystyle{\lim_{x \rightarrow 3}{ \left[ \frac{5x^2-8x-13}{x^2-5} \right] } }$$

Practice A04

$$\displaystyle{ \lim_{x \to 2}{ \frac{x^2-5x+6}{x^2-4} } }$$

##### Practice Problem [ A01 ] Solution

First of all, you always want to try direct substitution. When you do this, you get $$0/0$$ which is indeterminate.
There are at least two ways to solve this. Since this is a ratio of polynomials, we will use algebra. You can also use L'Hôpital's Rule.
First, we factor the numerator and denominator and cancel like terms, as follows:

$$\displaystyle{\lim_{x \rightarrow 2} \left[ \frac{3x^2-x-10}{x^2-4} \right] = \lim_{x \rightarrow 2} \left[ \frac{(x-2)(3x+5)}{(x-2)(x+2)} \right]}$$

Now the $$(x-2)$$ term that appears in both the numerator and denominator can cancel, giving

$$\displaystyle{\lim_{x \rightarrow 2} \left[ \frac{3x+5}{x+2} \right]}$$

Now let's try substitution again.

$$\displaystyle{\lim_{x \rightarrow 2} \left[ \frac{3x+5}{x+2} \right] = \frac{3(2)+5}{2+2} = 11/4 }$$

This technique will work only with polynomials that can be factored.

$$\displaystyle{\lim_{x \rightarrow 2}{ \left[ \frac{3x^2-x-10}{x^2-4} \right] }= 11/4 }$$

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##### Practice Problem [ A02 ] Solution

First of all, you always want to try direct substitution. When you do this in this limit, you get $$0/0$$ which is indeterminate. (If you don't know why this is, reread the indeterminate forms section.)
There are at least two ways to solve this. We will work it using algebra (since this is a ratio of polynomials). ( You can also use L'Hôpital's Rule, since direct substitution yields an indeterminate form ).
First, we factor the numerator and denominator and cancel like terms.
$$\displaystyle{ \lim_{x \rightarrow 3} \left[ \frac{x^4-81}{2x^2-5x-3} \right] = \lim_{x \rightarrow 3} \left[ \frac{(x^2-9)(x^2+9)}{(x-3)(2x+1)} \right] = \lim_{x \rightarrow 3} \left[ \frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)} \right] }$$
Now the $$(x-3)$$ term that appears in both the numerator and denominator can cancel, giving
$$\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{(x+3)(x^2+9)}{(2x+1)} \right] }$$
Now let's try substitution again.
$$\displaystyle{ \lim_{x \rightarrow 3} \left[ \frac{(x+3)(x^2+9)}{(2x+1)} \right] = \frac{(3+3)(3^2+9)}{(2(3)+1)} = 108/7}$$.
This way will work with only polynomials and only if you can factor.

$$\displaystyle{ \lim_{x \rightarrow 3}{ \left[\frac{x^4-81}{2x^2-5x-3}\right] }= 108/7 }$$

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##### Practice Problem [ A03 ] Solution

You always want to try direct substitution first.
$$\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{5x^2-8x-13}{x^2-5} \right] = \frac{5(9)-8(3)-13}{9-5} = \frac{45-24-13}{4} = \frac{8}{4} = 2 }$$
In this case, the result is determinate. So the limit is 2.

$$\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{5x^2-8x-13}{x^2-5} \right] = 2 }$$

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##### Practice Problem [ A04 ] Solution

$$\displaystyle{ \lim_{x \to 2}{ \frac{x^2-5x+6}{x^2-4} } = -1/4 }$$

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Practice A05

$$\displaystyle{ \lim_{x \to 1}{ \frac{x^2+x-2}{x^2-4x+3} } }$$

Practice A06

$$\displaystyle{ \lim_{x \to -1}{ \frac{x+1}{x^2-x-2} } }$$

Practice A07

$$\displaystyle{ \lim_{t \to -3}{ \frac{t^2+6t+9}{t^2-9} } }$$

Practice A08

$$\displaystyle{ \lim_{z \to -2}{ \frac{(z+2)^2}{z^4-16} } }$$

##### Practice Problem [ A05 ] Solution

She made a mistake in notation here. In the last column, she wrote $$\displaystyle{ \lim_{x \to 1}{ \frac{x+2}{x-3}} }$$ which is correct. However, in the first line of the next column, she should have written $$\displaystyle{ \frac{1+2}{1-3} = \frac{3}{-2} }$$ WITHOUT the limit as x goes to 1 in front of it. Once you substitute the value, you do not write the limit.

$$\displaystyle{ \lim_{x \to 1}{ \frac{x^2+x-2}{x^2-4x+3} } = -3/2 }$$

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##### Practice Problem [ A06 ] Solution

$$\displaystyle{ \lim_{x \to -1}{ \frac{x+1}{x^2-x-2} } = -1/3 }$$

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##### Practice Problem [ A07 ] Solution

$$\displaystyle{ \lim_{t \to -3}{ \frac{t^2+6t+9}{t^2-9} } = 0}$$

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##### Practice Problem [ A08 ] Solution

$$\displaystyle{ \lim_{z \to -2}{ \frac{(z+2)^2}{z^4-16} } = 0 }$$

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Practice A09

$$\displaystyle{ \lim_{x \to 1}{\frac{3x+5}{x+4} } }$$

Practice A10

$$\displaystyle{\lim_{x \to 0}{ ( x^2+x-6 ) } }$$

Practice A11

$$\displaystyle{ \lim_{x \to 14}{ (x+\sqrt{x+11} ) } }$$

Practice A12

$$\displaystyle{ \lim_{x \to 2}{\frac{x^2-4}{x-2}} }$$

##### Practice Problem [ A09 ] Solution

$$\displaystyle{ \lim_{x \to 1}{\frac{3x+5}{x+4} = 8/5} }$$

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##### Practice Problem [ A10 ] Solution

$$\displaystyle{ \lim_{x \to 0}{ ( x^2+x-6 ) = -6} }$$

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##### Practice Problem [ A11 ] Solution

$$\displaystyle{ \lim_{x \to 14}{ (x+\sqrt{x+11} ) = 19 } }$$

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##### Practice Problem [ A12 ] Solution

$$\displaystyle{ \lim_{x \to 2}{\frac{x^2-4}{x-2}} = 4 }$$

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Practice A13

$$\displaystyle{ \lim_{h \to 0}{ \frac{(4+h)^2-16}{h} } }$$

Practice A14

$$\displaystyle{ \lim_{z \to 2}{\frac{z^2+2z-8}{z^4-16}} }$$

Practice A15

$$\displaystyle{ \lim_{x \to -1}{ \frac{2x+2}{x+1}} }$$

Practice A16

$$\displaystyle{ \lim_{x \to 3}{\frac{x^2-6x+9}{x^2-9}} }$$

##### Practice Problem [ A13 ] Solution

$$\displaystyle{ \lim_{h \to 0}{ \frac{(4+h)^2-16}{h} = 8 } }$$

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##### Practice Problem [ A14 ] Solution

$$\displaystyle{ \lim_{z \to 2}{\frac{z^2+2z-8}{z^4-16}} = 3/16 }$$

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##### Practice Problem [ A15 ] Solution

$$\displaystyle{ \lim_{x \to -1}{ \frac{2x+2}{x+1}} = 2 }$$

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##### Practice Problem [ A16 ] Solution

$$\displaystyle{ \lim_{x \to 3}{\frac{x^2-6x+9}{x^2-9}} = 0 }$$

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Practice A17

$$\displaystyle{ \lim_{x \to 1}{\frac{x^2+x-2}{x-1}} }$$

##### Practice Problem [ A17 ] Solution

$$\displaystyle{ \lim_{x \to 1}{\frac{x^2+x-2}{x-1}} = 3 }$$

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 Level B - Intermediate

Practice B01

$$\displaystyle{ \lim_{t \rightarrow -2}{ \left[ \frac{ \frac{1}{t} + \frac{1}{2}}{t^3+8} \right] } }$$

Practice B02

$$\displaystyle{ \lim_{y \rightarrow 4}{ \left[ \frac{3-\sqrt{y+5}}{y-4} \right] } }$$

Practice B03

$$\displaystyle{\lim_{x \rightarrow 0}{ \left[ \frac{x^3-7x}{x^3} \right] } }$$

Practice B04

$$\displaystyle{ \lim_{x \rightarrow 0}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]} }$$

##### Practice Problem [ B01 ] Solution

The first thing you want to try here is to plug in -2 for t. However, in this limit, we get 0/0, which is an indeterminate form. So let's try factoring. But first we need to get the fraction in a different form.
In the numerator, we have $$\displaystyle{ \frac{1}{t} + \frac{1}{2} = \frac{2}{2t} + \frac{t}{2t} = \frac{t+2}{2t} }$$.

So the limit looks like $$\displaystyle{\lim_{t \rightarrow -2}{ \left[ \frac{ \frac{t+2}{2t}}{t^3+8} \right]} = \lim_{t \rightarrow -2}{ \left[ \frac{t+2}{(2t)(t^3+8)} \right]} }$$.

Now we need to factor the denominator. Let's check quickly to make sure (t+2) is a factor. This is a great time to practice synthetic division. Using synthetic division, $$t^3+8$$ factors into $$(t+2)(t^2-2t+4)$$.
So our limit is now $$\displaystyle{\lim_{t \rightarrow -2}{ \left[ \frac{t+2}{(2t)(t+2)(t^2-2t+4)} \right]} }$$. We can cancel the (t+2) term in the numerator and denominator and get $$\displaystyle{\lim_{t \rightarrow -2} \left[ \frac{1}{(2t)(t^2-2t+4)} \right]}$$.
Substituting -2 for t gives us $$\displaystyle{ \frac{1}{(-4)((-2)^2-2(-2)+4)} = \frac{1}{-4(4+4+4)} = -1/48 }$$

$$\displaystyle{ \lim_{t \rightarrow -2}{ \left[ \frac{ \frac{1}{t} + \frac{1}{2}}{t^3+8} \right] }= -1/48 }$$

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##### Practice Problem [ B02 ] Solution

As usual, you want to plug in 4 for y and see what you get. In this limit, you get 0/0, which is indeterminate. Our next idea might be to factor. However, there doesn't appear to be anything we can factor here. So, we have to do something called rationalizing.
To do this, we take the term with the square root in it and multiply the numerator and denominator by its conjugate. In this case, the conjugate of $$3-\sqrt{y+5}$$ is $$3+\sqrt{y+5}$$. The conjugate is obtained by changing the sign between the terms.
So the limit becomes
$$\displaystyle{ \lim_{y \rightarrow 4}{ \frac{3-\sqrt{y+5}}{y-4} } = \lim_{y \rightarrow 4}{ \left[ \frac{3-\sqrt{y+5}}{y-4} \right] \left[ \frac{3+\sqrt{y+5}}{3+\sqrt{y+5}} \right] } }$$
Keep in mind that we need to multiply the numerator and denominator by the SAME EXPRESSION. In essence, this means we are multiplying by 1. Otherwise we change the problem (which we don't want to do).
Okay, so now what do we do? Now we multiply out the numerator and leave the denominator factored (you will see why in a minute).
The numerator is $$\displaystyle{ (3-\sqrt{y+5})(3+\sqrt{y+5}) = 9 - 3\sqrt{y+5} + 3\sqrt{y+5} - (y+5) }$$. Do you see where all those terms came from?
Now, notice that there are two terms that look like $$3\sqrt{y+5}$$ with different signs. So they cancel each other. So we are left with $$9-(y+5) = 9-y-5 = 4-y$$. That's what the numerator reduces to. Our limit now looks like this.
$$\displaystyle{ \lim_{y \rightarrow 4}{ \left[ \frac{4-y}{(y-4)(3+\sqrt{y+5})} \right] } }$$
We are almost there. It looks like we have a term to cancel in the numerator and denominator but something is different. One of them is $$4-y$$ and the other is $$y-4$$. So they are slightly different. We need to do one more thing before we can cancel (the terms have to be EXACTLY the same to cancel). We need to get $$4-y$$ to be $$y-4$$. Well, what if we just switch the terms. If we do that we get $$-y+4$$ and that is obviously not $$y-4$$. (Remember, the sign in front of each term 'sticks' to that term, so when we switch the terms, we need to keep the negative sign with the y.)
To change $$(-y+4)$$ to $$(y-4)$$ we need to factor out a -1; (-y+4) = -1(y-4) giving us
$$\displaystyle{ \lim_{y \rightarrow 4}{ \left[ \frac{-1(y-4)}{(y-4)(3+\sqrt{y+5})} \right] } }$$.
Now you can see why we didn't want to multiply out the denominator. If we had, it would not have been obvious that we can cancel the $$(y-4)$$ in numerator and denominator. Doing so we get
$$\displaystyle{ \lim_{y \rightarrow 4}{ \left[ \frac{-1}{(3+\sqrt{y+5})} \right] } }$$ and substituting 4 for y gives us $$\displaystyle{ \frac{-1}{3+\sqrt{4+5}} = -1/6 }$$.

Note:
This technique works whether the square root term is in the numerator or denominator and it doesn't matter if the square root term appears first or second. Just change one of the signs and not the other.

$$\displaystyle{ \lim_{y \rightarrow 4}{ \left[ \frac{3-\sqrt{y+5}}{y-4} \right] } = -1/6 }$$

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##### Practice Problem [ B03 ] Solution

$$\displaystyle{\lim_{x \rightarrow 0}{\left[ \frac{x^3-7x}{x^3} \right]} = \lim_{x \rightarrow 0}{\left[ \frac{x(x^2-7)}{x(x^2)} \right]} = \lim_{x \rightarrow 0}{\left[ \frac{x^2-7}{x^2} \right]} }$$
The trick here is determine if the limit is $$+\infty$$, $$-\infty$$ or DNE (does not exist). So we need to look at the limit as x approaches zero from the left and compare it to the limit from the right. If they are different, then the limit does not exist.
Let's look at the left side of zero first. When x is less than zero (negative), $$x^2$$ is always positive. So the numerator is negative and the denominator is positive telling us that $$\displaystyle{ \lim_{x \rightarrow 0^- }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }$$.
Now, looking at the right side of zero, x is positive and $$x^2$$ is also positive. Since the numerator is negative and the denominator is positive, we know that $$\displaystyle{ \lim_{x \rightarrow 0^+ }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }$$.
Since $$\displaystyle{ \lim_{x \rightarrow 0^- }{\left[ \frac{x^2-7}{x^2} \right]}= \lim_{x \rightarrow 0^- }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }$$, the limit $$\displaystyle{ \lim_{x \rightarrow 0 }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }$$

$$\displaystyle{\lim_{x \rightarrow 0}{ \left[ \frac{x^3-7x}{x^3} \right] } = -\infty }$$

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##### Practice Problem [ B04 ] Solution

Direct substitution yields $$\displaystyle{ \lim_{x \rightarrow 0}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]} = \frac{-3}{0}}$$. So the answer is $$+\infty$$, $$-\infty$$ or DNE (Does Not Exist). How do you know which one (without a graph)?
We need to determine the behavior of the function on either side of $$x=0$$. On the left side of zero when $$x < 0$$ and is very close to zero, the numerator is negative. We know this since x is very, very small, so $$x^4$$ is positive but very small and $$5x$$ is very small but negative. Both terms will be dominated by -3, so the numerator is negative. In the denominator, we need to look at the square root; $$x^2$$ is always positive and adding a positive number to 4 means that the number under the square root is always greater than 4, so the square root is always larger than 2. This makes the denominator negative. So we have a negative numerator and a negative denominator, so the limit $$\displaystyle{ \lim_{x \rightarrow 0^-}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}= +\infty }$$
We use similar logic when x > 0, to get $$\displaystyle{ \lim_{x \rightarrow 0^+}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}= +\infty }$$

Since the limit from the right is equal to the limit from the left, the original limit is also $$+\infty$$.

$$\displaystyle{ \lim_{x \rightarrow 0}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}= +\infty }$$

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Practice B05

$$\displaystyle{ \lim_{z \rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]} }$$

Practice B06

$$\displaystyle{ \lim_{x \to 1}{ \frac{x^3-1}{x^4-1} } }$$

Practice B07

$$\displaystyle{ \lim_{h \to 0}{ \frac{(3+h)^{-1}-3^{-1}}{h} } }$$

##### Practice Problem [ B05 ] Solution

Direct substitution yields $$0/0$$ which is indeterminate.
Factoring yields
$$\displaystyle{ \lim_{z \rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]} = \lim_{z \rightarrow 1}{\left[\frac{(z-1)(z^2+z+1)}{(z-1)^2}\right]} = \lim_{z \rightarrow 1}{\left[\frac{z^2+z+1}{z-1}\right]}}$$
Direct substitution here gives us $$3/0$$. So is the answer $$+\infty$$ or $$-\infty$$? (without looking at a graph)
Well, looking at values close to $$1$$, when $$z < 1$$ the result is negative and when $$z>1$$ the result is positive. So the answer is neither $$+\infty$$ nor $$-\infty$$ and the only correct is answer is that the limit does not exist.

$$\displaystyle{ \lim_{z \rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]} }$$ does not exist.

Notes
See the section on Limits That Do Not Exist for more information.

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##### Practice Problem [ B06 ] Solution

$$\displaystyle{ \lim_{x \to 1}{ \frac{x^3-1}{x^4-1} } = 3/4 }$$

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##### Practice Problem [ B07 ] Solution

$$\displaystyle{ \lim_{h \to 0}{ \frac{(3+h)^{-1}-3^{-1}}{h} } = -1/9 }$$

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Practice C01

$$\displaystyle{ \lim_{x \rightarrow 27}{\left[ \frac{x - 27}{x^{1/3} - 3}\right]} }$$

Practice C02

$$\displaystyle{ \lim_{x \rightarrow 1}{\left[ \frac{x^{1/3}-1}{x^{1/4}-1}\right] } }$$

##### Practice Problem [ C01 ] Solution

Of course, the first thing we want to try is to plug in 27 directly for x. This gives us $$0/0$$ which is indeterminate. So we need to do something else.
Upon initial inspection, this expression doesn't seem to be factorable. However, I am going to show you a trick. I don't know about you but I am not completely comfortable working with fractional powers. So I am going substitute $$t$$ for $$x^{1/3}$$. We can do that as long we do the substitution for every expression containing $$x$$. So what do we get?

$$\displaystyle{ \frac{x-27}{x^{1/3}-3} = \frac{t^3-27}{t-3} }$$

Notice that since $$t=x^{1/3}$$, we can cube both sides to get $$t^3=x$$.

Using synthetic division, we can factor the numerator which becomes $$(t^3-27)=(t-3)(t^2+3t+9)$$.

Now the expression is $$\displaystyle{ \frac{(t-3)(t^2+3t+9)}{t-3}}$$ and we can easily see that the $$t-3$$ will cancel.
So let's go back to x's since our limit is in terms of x.

$$\displaystyle{ \begin{array}{rcl} \lim_{x \rightarrow 27}{\frac{x-27}{x^{1/3}-3}} & = & \lim_{x \rightarrow 27}{\frac{(x^{1/3}-3)(x^{2/3}+3x^{1/3}+9)}{x^{1/3}-3}} \\ & = & \lim_{x \rightarrow 27}{\frac{(x^{2/3}+3x^{1/3}+9)}{1}} \\ & = & 27^{2/3}+3(27)^{1/3}+9 = 27 \end{array} }$$

$$\displaystyle{ \lim_{x \rightarrow 27}{\left[ \frac{x - 27}{x^{1/3} - 3}\right]} = 27 }$$

Notes>
We could have used L'Hôpital's Rule here, which would have been much easier.
More information on the technique of substitution - 17Calculus: Technique Of Substitution
Review synthetic division here: 17Calculus: Synthetic Division

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##### Practice Problem [ C02 ] Solution

Direct substitution yields $$0/0$$, which is indeterminate. To solve this exercise we need to do a trick.
We want to see if we can factor this problem to cancel terms in the numerator and denominator. I don't know about you but I have a hard time working with fractional powers. So I am going to use substitution to remove fractional powers.
In this problem, I am going to substitute for x. I want to get rid of both the 1/3rd and 1/4th powers. So, to do that I am going to let $$t=x^{1/12}$$. I chose 1/12 since 1/12=(1/3)(1/4). You will see in a minute why this works.
Okay, so we have $$t=x^{1/12}$$ which means $$t^{12}=x$$ (by taking the 12th power of both sides).
Now, let's substitute.
$$\displaystyle{ \frac{x^{1/3}-1}{x^{1/4}-1} = \frac{(t^{12})^{1/3}-1}{(t^{12})^{1/4}-1} = \frac{t^4-1}{t^3-1} }$$
Okay, so I can factor the numerator and denominator here (use synthetic division) and I get
$$\displaystyle{ \frac{(t-1)(t^3+t^2+t+1)}{(t-1)(t^2+t+1)}}$$
Even though this is not completely factored, we can see that we can cancel (t-1) in this fraction, so we can stop factoring here.
Okay, let's go back to limit notation and convert back to x's.
$$\displaystyle{ \lim_{x \rightarrow 1}{\frac{(x^{1/12}-1)(x^{1/4}+x^{1/6}+x^{1/12}+1)}{(x^{1/12}-1)(x^{1/6}+x^{1/12}+1)}} = \lim_{x \rightarrow 1}{\frac{x^{1/4}+x^{1/6}+x^{1/12}+1}{x^{1/6}+x^{1/12}+1}}}$$
Now when we substitute 1 for x, we get $$\displaystyle{ \frac{1+1+1+1}{1+1+1} = 4/3 }$$

$$\displaystyle{ \lim_{x \rightarrow 1}{\left[ \frac{x^{1/3}-1}{x^{1/4}-1}\right] }= 4/3 }$$